Question

The value of $\frac{3\times 1^3}{1^2}+\frac{5\times (1^3+2^3)}{1^2+2^2}+\frac{7\times (1^3+2^3+3^3)}{1^2+2^2+3^2}+\cdots \text{upto}10\text{terms}$, is

• A. $600$
• B. $620$
• C. $660$
• D. $680$

Answer 1

The correct option is C $660$

Let
$S=\frac{3\times 1^3}{1^2}+\frac{5\times (1^3+2^3)}{1^2+2^2}+\frac{7\times (1^3+2^3+3^3)}{1^2+2^2+3^2}+\cdots \text{upto}10\text{terms}$

So, the general term is
$T_n=\frac{(2n+1)(1^3+2^3+3^3+\cdots +n^3)}{(1^2+2^2+3^2+\cdots +n^2)}=(2n+1)\frac{{\left(\frac{n(n+1)}{2}\right)}^2}{\frac{n(n+1)(2n+1)}{6}}$
$=\frac{3n(n+1)}{2}$

Now,
$S=\sum _{n=1}^{10}{T}_n=\frac{3}{2}[\sum _{n=1}^{10}{n}^2+\sum _{n=1}^{10}n]=\frac{3}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]=\frac{3}{2}[\frac{10\cdot \left(11\right)\cdot \left(21\right)}{6}+\frac{10\cdot \left(11\right)}{2}]\therefore S=660$